Grams ethanol = 33 ml times .789 gms/ml = 26.037 gms
<span>Moles ethanol = 26.037 gms / 46 gms/mole = .57 moles </span>
<span>Moles water = 67 ml or 67 grams/18 gms/mole = 3.22 moles </span>
<span>total moles = .57 + 3.72 = 4.29 moles </span>
<span>Mole fraction ethanol = .57 moles ethanol / 4.29 moles total = 0.13</span>
<span>Moles fraction water = 3.72 moles water / 4.29 moles total = 0.87</span>
<span>Partial pressure of ethanol = mole fraction ethanol (.13) _ times VP ethanol 43.9 torr) = 5.707 torr </span>
<span>partial pressure water = mole fraction water .87) times VP water (l7.5 torr) = 15.23 torr </span>
<span>Total vapor pressure over solution = 5.71 torr + 15.23 torr = 20.94 torr</span>
Answer:
It’s b even check quizzes
Explanation:
It seems that you have missed to attach the given image for us to answer this question, but anyway, here is the answer. Based on the picture of a box which is motionless, the resulting motion of the box would be option B. The box will move to the left. Hope this answer helps.
Answer:
(a) The proportion of dry air bypassing the unit is 14.3%.
(b) The mass of water removed is 1.2 kg per 100 kg of dry air.
Explanation:
We can express the proportion of air that goes trough the air conditioning unit as 
and the proportion of air that is by-passed as 
, being 
.
The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:
Replacing the first equation in the second one we have
(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.
It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.
The water removed of every 100 kg of dry air is
It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).
100 * 0.012 = 1.2 kgW
Answer:
Mn2O3
Explanation:
Manga has a 3+ charge and oxygen has a 2- charge so to balance the charges there needs to be 3 oxygens for every 2 manga
